/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Find the Maclaurin series for th... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the Maclaurin series for the functions \(x^{4}-2 x^{3}-5 x+4\)

Short Answer

Expert verified
The Maclaurin series is \( 4 - 5x - 2x^3 + x^4 \).

Step by step solution

01

Understand the Maclaurin Series

A Maclaurin series is a Taylor series expansion of a function about 0, given by: \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). For polynomial functions, the series becomes finite because higher-order derivatives beyond the degree of the polynomial will be zero.
02

Find the Derivatives

To find the Maclaurin series expansion of \( f(x) = x^{4}-2x^{3}-5x+4 \), we calculate the derivatives at \( x = 0 \). First derivative: \( f'(x) = 4x^3 - 6x^2 - 5 \) Second derivative: \( f''(x) = 12x^2 - 12x \) Third derivative: \( f'''(x) = 24x - 12 \) Fourth derivative: \( f^{(4)}(x) = 24 \) Fifth derivative and beyond will be zero because the polynomial only has degree 4.
03

Evaluate Derivatives at 0

Now evaluate each of these derivatives at \( x=0 \): \( f(0) = 4 \)\( f'(0) = -5 \)\( f''(0) = 0 \)\( f'''(0) = -12 \)\( f^{(4)}(0) = 24 \).
04

Construct the Series

Substitute these values into the Maclaurin series formula: \[ f(x) = 4 + (-5)x + \frac{0}{2!}x^2 + \frac{-12}{3!}x^3 + \frac{24}{4!}x^4 \]Simplify this to:\[ f(x) = 4 - 5x - 2x^3 + x^4 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series
A Taylor Series is a powerful mathematical tool that allows us to approximate functions using an infinite sum of terms calculated from the function's derivatives at a single point. In general, a Taylor Series is defined around a point "a" as:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \)
In the specific case of a Maclaurin Series, "a" is zero, thus the expansion becomes:
  • \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \)
This series helps in expressing complex functions in terms of more straightforward polynomial expressions, making them easier to analyze and compute.
Polynomial
A polynomial is a mathematical expression consisting of variables, coefficients, and exponents. In general, a polynomial is written as:
  • \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)
Polynomials are characterized by their degree, which is the highest power of the variable in the expression. For example, in the polynomial \( x^4 - 2x^3 - 5x + 4 \), the degree is four. These expressions are often straightforward to differentiate and integrate, and are fundamental to algebra and calculus.The reason why a polynomial such as \( x^4 - 2x^3 - 5x + 4 \) results in a finite Maclaurin Series is because the derivatives beyond its highest degree—fourth in this case—are zero. This simplifies calculations as no terms exist beyond those initial derivatives.
Derivative
The derivative of a function is a central concept in calculus that measures how a function changes as its input changes. It provides the rate of change or the slope of the function at a particular point. The derivative of a function \( f(x) \) with respect to \( x \) is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).Calculating derivatives is essential for constructing Taylor or Maclaurin series. It involves successively finding the first, second, third derivatives, and so on. Take, for instance, the polynomial function \( f(x) = x^4 - 2x^3 - 5x + 4 \):
  • The first derivative, \( f'(x) = 4x^3 - 6x^2 - 5 \), shows how the polynomial changes per unit of \( x \).
  • The second derivative, \( f''(x) = 12x^2 - 12x \), provides information on the concavity of the graph.
  • Third derivative, \( f'''(x) = 24x - 12 \), gives insights into how the rate of change itself is changing.
  • The fourth derivative, \( f^{(4)}(x) = 24 \), indicates changes at a higher level of calculus insight but for polynomials of degree 4, these are the limits of necessary derivatives.
The derivatives evaluated at \( x = 0 \) are used to build the Taylor and Maclaurin series, highlighting how derivatives are fundamental to these mathematical representations.

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Most popular questions from this chapter

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{1}{1+2^{2}+3^{2}+\cdots+n^{2}}\end{equation}

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Logarithmic \(p\) -series a. Show that the improper integral $$\int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant })$$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

How many terms of the convergent series \(\sum_{n=4}^{\infty}\left(1 / n(\ln n)^{3}\right)\) should be used to estimate its value with error at most 0.01\(?\)

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2 .\) Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

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